Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 14 / a

Distinguish the production type as well as the stopping position. The sample space has the six outcomesHere $x_k$ in an event means $X=x_k$. Their joint probabilities are $p{g}_k$ and $(1-p)d_k$, respectively. The conditional stopping laws obey ${\sum }_{k=0}^2{g}_k={\sum }_{k=0}^2{d}_k=1$; with the stated strict positivity all six outcomes are possible.

Multiply the production probability by the conditional stopping probability:

The good and defective cases are disjoint and exhaustive. The law of total probability therefore gives

Divide the relevant joint probability by the marginal stopping probability. By Bayes' theorem,The denominator is positive under the stated assumptions.

The condition $X>x_0$ selects positions $x_1$ and $x_2$. Their combined good-ball joint probability is $p(g_1+g_2)$, and their total probability is $p(g_1+g_2)+(1-p)(d_1+d_2)$. Thus